POJ-2065 SETI Gaussian elimination, extended GCD

This question is given as follows:

F (1) = C1 (MOD) P
F (2) = c2 (MOD) P
F (3) = C3 (MOD) p...

F (1) = a () * X1 + a (1, 2) * X2 + a (1, 3) * x3...

Where a (I, j) = I ^ (J-1 ).

In the face of such an equation, our approach is to ignore the (MOD) P on the right side of the equation, because f (I) and CI are the same, therefore, multiplying a number at both ends of an equation does not affect the performance of a given equation. So the rest is the Gaussian elimination element. Because the solution is unique, the obtained coefficient matrix must be full rank. Next, you can use the extended GCD to solve unknown elements.

The Code is as follows:

# Include <cstdlib> # include <cstring> # include <algorithm> # include <cstdio> using namespace STD; typedef long int int64; char s [100]; int length, MOD; int64 X [100]; int _ POW (int A, int B) {// fast power int ret = 1; while (B) {If (B & 1) {RET * = A; RET % = MOD;} A * = A; A % = MOD; B >>= 1;} return ret ;} struct matrix {int64 A [100] [100]; void Init () {for (INT I = 1; I <= length; ++ I) {for (Int J = 1; j <= Length; ++ J) {A [I] [J] = _ POW (I, J-1 );} A [I] [Length + 1] = s [I] ;}} void rswap (INT X, int y, int S) {// used to exchange the following rows to the first row for int64 t; for (Int J = s; j <= Length + 1; ++ J) {T = A [x] [J]; A [x] [J] = A [y] [J]; A [y] [J] = T ;}} void multi (int x, int64 M, int s) {// multiply the same row by a certain number. Of course, for (Int J = s; j <= Length + 1; ++ J) {A [x] [J] * = m;} void relax (int x, int y, int S) {// iterate an equation into another expression for (Int J = S; j <= Length + 1; ++ J) {A [y] [J]-= A [x] [J]; A [y] [J] = (a [y] [J] % mod + mod) % MOD; A [x] [J] = (a [x] [J] % mod + mod) % mod ;}} void print () {for (INT I = 1; I <= length; ++ I) {for (Int J = 1; j <= Length + 1; ++ J) {printf ("% 5i64d ", A [I] [J]);} puts ("");} puts ("") ;}} m; int64 gcd (int64 A, int64 B) {int64 T; while (B) {T = B; B = A % B; A = T;} return a;} int64 lcm (int64 A, int64 B) {return /Gcd (a, B) * B;} int64 ext_gcd (int64 A, int64 B, int64 & X, int64 & Y) {int64 temp, RET; If (! B) {x = 1, y = 0; return a;} ret = ext_gcd (B, A % B, x, y); temp = x, x = Y, y = temp-A/B * Y; return ret;} void Gauss () {int I = 1, K; int64 a, B, sum; memset (x, 0, sizeof (x); For (Int J = 1; j <= length; ++ J) {for (k = I; k <= length; ++ K) {If (M. A [k] [J]) break;} If (k> length) continue; If (K! = I) {M. rswap (I, K, j) ;}for (k = I + 1; k <= length; ++ K) {If (M. A [k] [J]) {int64 LCM = lcm (M. A [I] [J], M. A [k] [J]); A = LCM/m. A [I] [J]; B = LCM/m. A [k] [J]; if (a> 1) M. multi (I, A, J); If (B> 1) M. multi (K, B, J); M. relax (I, K, j) ;}++ I ;}// M. print (); // there must be a unique solution, so no undefined determination for (k = I-1; k> = 1; -- k) {// Of course, here can be brute force to solve, after all, the given prime number is not very large int64 XX, YY, a, B, c = 0, L, G; For (Int J = k + 1; j <= length; ++ J) {C ++ = x [J] * m. A [k] [J];} A = m. A [k] [K], B = MOD; C = m. A [k] [Length + 1]-C; G = ext_gcd (a, B, XX, YY); L = B/g; XX * = C/g; xx = (XX % L + l) % L; X [k] = XX ;}for (Int J = 1; j <= length; ++ J) {printf (j = 1? "% I64d": "% i64d", X [J]);} puts ("") ;}int main () {int N; scanf ("% d ", & N); While (n --) {scanf ("% d % s", & mod, S + 1); length = strlen (S + 1 ); for (INT I = 1; I <= length; ++ I) {If (s [I]! = '*') {S [I]-= ('A'-1) ;}else {s [I] = 0 ;}} M. init (); Gauss ();} return 0 ;}